What uses do products of power series have?

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What uses do products of power series have?

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Answer 1 · 2015-08-23T08:21:10

One example that I find useful is the use and manipulation of the products of power series to derive $e^{i x} = cos x + i sin x$ $e^(ix) = cosx + isinx$ , which is an identity used many, many times to solve the Schroedinger Equation in Physical Chemistry, by substituting $i$ $i$ for various different constants.

What is accepted by Physical Chemists is that you can write out the general solution to the Equation as:

$y (x) = c_{1} e^{α x} + c_{2} e^{- α x}$ $y(x) = c_1 e^(alphax) + c_2 e^(-alphax)$

where $α = i ω t$ $alpha = iomegat$ and, after various proofs and empirical tests, it is agreed that we can use $e^{α x}$ $e^(alphax)$ as a working "trial function" when we guess the form of the overall solution in terms of a finite addition of these $c \cdot e^{α x}$ $c*e^(alphax)$ functions so that we can predict molecular properties:

$ψ (x) = N \sum i = 1 c_{i} ϕ_{i} (x)$ $psi(x) = sum_(i=1)^N c_i phi_i(x)$
where each $ϕ$ $phi$ could, for example, represent an atomic orbital, and $ψ (x)$ $psi(x)$ would in that case be the molecular orbital.

A common example of solving the time-dependent Schroedinger equation is (example 2-4 in Physical Chemistry: A Molecular Approach):

$\frac{d^{2} x (t)}{{d t}^{2}} + ω^{2} x (t) = 0$ $(d^2x(t))/(dt^2) + omega^2x(t) = 0$

subject to the boundary conditions $x (0) = A$ $x(0) = A$ and $\frac{d x (0)}{d t} = 0$ $(dx(0))/(dt) = 0$ . These boundary conditions define the fact that a stationary transverse wave with one antinode has two endpoints, and these are at $x = 0$ $x = 0$ and $x = l$ $x = l$ , half of the wavelength.

![http://www.physicsclassroom.com/]( useruploads.socratic.org )

To solve this one, one would have to use identity written at the top, with $α$ $alpha$ substituted for $i$ $i$ like so:

$c_{1} e^{α x} + c_{2} e^{- α x}$ $c_1e^(alphax) + c_2e^(-alphax)$

$= c_{1} (cos x + α sin x) + c_{2} (cos x - α sin x)$ $= c_1(cosx + alphasinx) + c_2(cosx - alphasinx)$

$= c_{1} cos x + c_{1} α sin x + c_{2} cos x - c_{2} α sin x$ $= c_1cosx + c_1alphasinx + c_2cosx - c_2alphasinx$

$= (c_{1} + c_{2}) cos x + (c_{1} α - c_{2} α) sin x$ $= (c_1 + c_2)cosx + (c_1alpha - c_2alpha)sinx$

and it is generally written out by absorbing the arbitrary constants $c_{1}$ $c_1$ , $α$ $alpha$ , and $c_{2}$ $c_2$ into new arbitrary constants $c_{3}$ $c_3$ and $c_{4}$ $c_4$ , with $c_{1} + c_{2} = c_{3}$ $c_1 + c_2 = c_3$ and $c_{1} α - c_{2} α = c_{4}$ $c_1alpha - c_2alpha = c_4$ :

$= c_{3} cos x + c_{4} sin x$ $= c_3cosx + c_4sinx$

Then, substituting $ω t$ $omegat$ for $x$ $x$ , we get:

$c_{3} cos (ω t) + c_{4} sin (ω t)$ $c_3cos(omegat) + c_4sin(omegat)$

for the solution to the so-called common example.

Looking at the boundary condition $\frac{d x (0)}{d t} = 0$ $(dx(0))/(dt) = 0$ , we get:

$= c_{3} (- sin (ω t)) \cdot ω + c_{4} cos (ω t) \cdot ω$ $= c_3(-sin(omegat))*omega + c_4cos(omegat)*omega$

$= cancel(c_3(-sin(omega(0)))*omega)^(0) + c_4cos(omega(0))*omega$

$= c_4omega$

But we know that at $t = 0$ , $omega = 0$ because time has not passed yet.

$=> c_4omega = 0$ , thus satisfying the condition $(dx(0))/(dt) = 0$ .

Using the $x(0) = A$ boundary condition we get:

$x(0) = c_3cos(omega(0)) + cancel(c_4sin(omega(0)))^(0)$

$= c_3$

with $c_3$ taken as $A$ ---which is the amplitude of an initialized stationary wave---since it is the only contributor to the wave. Thus, since $c_3 = A$ , we have satisfied the condition $x(0) = A$ , and we just have:

$color(blue)(x(t) = Acos(wt))$

which is the familiar physics equation for a transverse wave, as depicted in the image above! :)

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What uses do products of power series have?

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